bzoj2081 Beads

枚举答案暴力就可以了，复杂度O(nlog2n)。

#include<cstdio>#include<algorithm>#include<set>#include<vector>using namespace std;#define LL unsigned long longconst int maxn=200010,p=13331;LL h[maxn],g[maxn],pow[maxn];int a[maxn],n,ans;set<LL> s;vector<int> v;int main(){    int cnt;    LL x,y;    scanf("%d",&n);    for (int i=1;i<=n;i++) scanf("%d",&a[i]);    for (int i=1;i<=n;i++) h[i]=h[i-1]*p+a[i];    for (int i=1;i<=n;i++) g[i]=g[i-1]*p+a[n-i+1];    pow[0]=1;    for (int i=1;i<=n;i++) pow[i]=pow[i-1]*p;    for (int i=1;i<=n;i++)    {        cnt=0;        s.clear();        for (int j=1;j+i-1<=n;j+=i)        {            x=h[j+i-1]-h[j-1]*pow[i];            y=g[n-j+1]-g[n-j-i+1]*pow[i];            x*=y;            if (!s.count(x)) s.insert(x),cnt++;        }        if (cnt>ans)        {            v.clear();            v.push_back(i);            ans=cnt;        }        else if (cnt==ans) v.push_back(i);    }    printf("%d %d\n",ans,(int)v.size());    for (int i=0;i<v.size();i++) printf("%d%c",v[i],i==v.size()-1?'\n':' ');}